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In his wardrobe, Dexter has three trousers. One of them is black the second is blue, and the third brown. In his wardrobe, he also has four shirts. One of them is black and the other 3 are white. He opens his wardrobe in the dark and picks out one shirt and one trouser pair without examining the colour. What is the likelihood that neither the shirt nor the trouser is black?









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Option(D) is correct

Probability that trouser is not black $=2/3$

Probability that shirt is not black $= 3/4$

Required probability $= 2/3 × 3/4 = 1/2$

(2) Comment(s)


Probability of choosing a black shirt = 1/4

Probability of choosing a black trouser = 1/3

So the probability of choosing a black shirt and a black trousers will be 1/4*1/3 = 1/12

therefore, probability of neither choosing a black shirt or a black trouser = 1 minus 1/12 = 11/12

What's wrong in this method ? Please help!


I think when you use this method you are calculating the probability of "black shirt and black trouser not coming together" as you subtracted the probability of both coming together and there will be combination where 1 black shirt + other trouser will come or vice versa, which would be wrong since in question it is asked that neither of them should come, so remove all such combinations. #Ashok